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<h1>Geometry</h1>

<p>The geometry takes into account that the 12th fret position divides the scale length into 2 halves of equal length. It is the position of the first harmonic of the open string frequency. </p>

<p>Question is, where exactly is the reference plane? a) The lower end of the string, because the string can be considered a straight line when under tension.</p>


<p>A formula by Lester Backshall, Guitar Maker, Aylesbury, UK (all values given in mm):</p>

<pre class="codebox formula">
SH = ( FB + FR + ( A * 2 )) - ( DH + N )
    with
    SH = Saddle Height (12)
    FB = Fingerboard thickness (6)
    FR = Fret Height (1)
    A = Action at F12 (3.5 @ E6)
    DH = Doming at bridge (2)
    N = Neck Set (?)
</pre>

<ul>
<li>Reference plane is the underside of the string.</li>
<li>Action is defined as spacing between fret top and string underside.</li>
<li>Action at zero fret (nut) (A0) is considered zero.</li>
<li>Scale length (SL) = 650 mm.</li>
</ul>

<p>With these assumptions we can calculate the angle between fretboard top and string:</p>

<pre classs="codebox formula">
tan a = 3.5 / 650 * 2 = 3.5 / 325
a = arctan ( 3.5 / 325 ) = 0.6170076 degrees (rounded)
</pre>

<p>If we extend the triangle, defined by the action of 3.5 mm at the 12th fret (A12), to the bridge, we can easily calculate the position where the plane defined by the top of the frets intersects the bridge (AB)</p>

<pre class="codebox formula">
tan a = AB / SL
AB = tan a * SL = tan a * 650 = 3.5 / 325 * 625 = 3.5 * 2 = 7
</pre>

<p>This is exactly what the rule of length relations for similar triangles gives, so we didn't acutally had to use trigonometry.</p>

<p>As we have our reference plane chosen to be the underside of the string, the effective height of the fretboard, seen from this plane, is a little bit greater than the measured fretboard thickness: We don't look perpendicular on the fretboard, but at a slight angle, and this angle is the angle a as calculated above. So the effective fretboard height (EFBH) is calculated as</p>

<pre class="codebox formula">
cos a = FB / EFBH
EFBH = FB / cos a = 6 / cos ( arctan ( 3.5 / 325 )) 
    = 6.000348 mm (rounded)
</pre>

<p>Provided, the fretboard not tapered lengthwise, and is 6 mm thick (FB), and the soundbord is perfectly flat, then the fretboard underside plane would intersect the bridge at ...</p>

<pre class="codebox formula">
AFU = AB + EFBH = 7 + 6.000348 = 13.000348 mm
</pre>

<p>...  measured perpendicular from the string underside. Because soundboard upperside plane is the same a the fretboard underside plane, we can say that this is the space between string and soundboard.</p>

<p>Now the doming of the soundboard comes into play. Radius and shape of the dome is irrelevant here, the only thing we have to know is that we carved a dip into the solera, with depth (DH) of 5 mm at the bridge, and that the area at fret 12, where the neck meets the body, has been left untouched. Then we can simply subtract that depth from the AFU height:</p>

<pre class="codebox formula">
AFU = AB + EFBH - DH = 7 + 6.000348 - 5 = 8.000348 mm
</pre>

<p>This is a little bit flat for a classical guitar. We have now several options:</p>

<ul>
<li>Raise the fretboard by putting some support underneath, or</li>
<li>angle the neck back slightly, so that underside of the fretboard points a little bit higher at the bridge position, or</li>
<li>bring the center of the dome radius more to 12th fret, away from the bridge.</li>
</ul>

<p>The second option might be easy to achive, because the dome comes to the 12th fret as a certain angle, wich is defined by radius and scale length. And if this angle matches the desired nect angle, the neck would touch the dome radius exactly tangentially &mdash; which means practically that we don't have to worry much about fretboard tapering in the area of the upper bout. </p>

<p>The third method, although it sounds different at first sight, introduces exactly the same angle as option 2, just the point of view is different. Instead of tilting the neck back slightly, we angle the body back sightly. The effect is the same. I guess, this is what Romanillos and Bogdanovich did with their approach.</p>

<p>All corrections have to go into the math. The first option is quite simple, it add just another offset (fretboard height support, FBHS) into the AFU formula:</p>

<pre class="codebox formula">
AFU = AB + EFBH - DH + FBHS = 7 + 6.000348 - 5 + 4 = 12.000348 mm
</pre>

<p>The second one is more complicated, because it changes the geometry. Or does it?</p>

<h2>General Discussion of Neck Angles</h2>

<p>Suppose the guitar top follows a section P1, P3 of a circle with the radius DR, centered at DC, including the angles DA, FDA and BDA. The neck is meets the top at P1, the upper end of the body, P2 is the position where the bridge goes on the top, and P3 is the lower end of the body. The line through P1&ndash;P3 is the body length BL, and represents a perfectly flat top, a solera blank before beeing carved. The distance P4&ndash;P2 is the dome height DH at the bridge position, the depth to which the solera is carved out under the bridge.</p>

<p>First let's assume the neck is tilted back quite far, so that the underside plane is perpendicular to the radius DC&ndash;P1. Then the underside of the fretboard falls on the tangent of the dome circle, which provides a smooth transition from the neck to the top surface. The neck angle NA is defined by the tangent through P1, and the line P1&ndash;P3.</p>

<p>Soon after the neck underside meets the top, both surfaces separate from each other, leaving a gap between fret board underside and soundboard above the sound hole. This is not what we want. Also, the soundboard points way to high in the air at the bridge position P2.</p>

<p>But what we can see from this model, is that a larger dome radius DR implies</p>

<ul>
<li>a smaller neck angle NA</li>
<li>a smaller bridge height</li>
<li>a largher home height DH</li>
</ul>

<p>Now let's lower the neck angle a little bit. This means that the fretboard underside crosses the top surface at P1, diving into the soundboard. Let's choose the angle so that the underside crosses the top surface again at the upper end of the sound hole, where the fretboard ends. The intersection of them is a tangential section of the dome circle, which is kind of a problem. We have two options:</p>

<ul>
<li>Carve the fretboard underside, or sand it down to match the soundboard profile.</li>
<li>Leave the fretboard flat, and clamp it down while glueing it to the top.</li>
</ul>

<p>Sanding is easy if you have a 27 foot disk on you grinder ;-) but you must get it hollow, because the profile is actually a section of a sphere. A combination of small konvex plane, gouge, and sandpaper might do the job. For final adjustments, one could stick a piece of sandpaper on the top and rub the fretboard end over it until it fits perfectly.</p>

<p>Another option would be to flatten the soundboard under the fretboard, but that is limited by the soundboard thickness in that area. A reinforcement patch on the soundboard underside might be necessary then, because it is reported that the soundboard tends to crack near the fretboard sides.</p>

<p>This might be a lot of work, so it's worth calculating the sphere section height in order to estimate the second option.</p>

<p>A first, rough estimation of the upper limit sets the dome height DH, the halfed scale length SL, and the distance from sound hole upper corner to upper body corner (assumed 105 mm) in a relation: </p>

<pre class="codebox formula">
SL / 2 / DH = 105 / x
x = 105 * 2 * DH / SL = 201 * 5 / 650 = 1.6 mm (rounded)
</pre>

<p>The real value should be a little bit smaller, perhaps 1.3 to 1.5 mm.</p>

<p>Exact calculation of section height can be calculated by using the Pythagorean theorem. We divide the 105 mm secant into halves, which gives us 2 orthogonal triangles P1 P2 PDC, and P3 P2 PDC. The hypothenusis length is the dome radius LDR, one leg is 52.5 mm, so the other leg is the square root of the difference of the squares. This subtracted from the dome radius is our desired dome height under the fretboard end. </p>

<pre class="codebox formula">
P1-P3 = 105
P1-P2 = P2-P3 = P1-P3 / 2 = 52.5
P1-PDC = P3-PDC = LDR = 7620
AP1-P2-PDC = 90 degree
squ ( P2-PDC) = squ ( P1-PDC ) - squ ( P1-P2 )
P2-PDC = sqrt ( squ ( P1-PDC ) - squ ( P1-P2 ))
x = LDR - P2-PDC = LDR - sqrt ( squ ( P1-PDC ) - squ ( P1-P2 ))
  = 7620 - sqrt ( squ ( 7620 ) - squ ( 52.5 ))
  = 7620 - sqrt ( ??? )
  = 7620 - ???
  = 0.18 mm (rounded)
</pre>

<p>That is so small that it is not worth to scratch the head. Just glue it.</p>

<p></p>


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